Complex number, power series

Question

Develop $\sinh z$ in powers of $z-\pi i$ to show that $$\lim_{z\to \pi i}\frac{\sinh z}{z-\pi i}=-1$$

I know that $\sinh z=\sum_{n=1}^\infty \frac{z^{2n-1}}{(2n-1)!}$.

Edit:

Following the hint I get that $\sinh{(z-\pi i)}=-\sinh{z}$ Then $$\sinh{(z-\pi i)}=-\sum_{n=1}^\infty \frac{z^{2n-1}}{(2n-1)!}$$

But I do not know how to use it to prove that limit, honestly I'm not understand the connection between the power series and the limit

No, that would be $\sinh(z-\pi i)$, which is something different. — Thomas Andrews, May 10, 2015 at 0:06
But you also said, "I suppose that..." and then stated an equality that I was contradicting. No reason to be defensive. — Thomas Andrews, May 10, 2015 at 0:08

Thomas Andrews · Accepted Answer · 2015-05-10 00:10:54Z

4

Hint: $\sinh(a+b)=\sinh a\cosh b+\cosh a\sinh b$. Then use $a=z-\pi i$ and $b=\pi i$.

answered May 10, 2015 at 0:10

Thomas Andrews

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$\begingroup$ I don't think that answers the question. The problem wants to use power series. $\endgroup$
– user223391
May 10, 2015 at 0:42
1

$\begingroup$ @avid19 That's why it is a hint - there is a step after this, using the power series for $\cosh w$ and $\sinh w$. $\endgroup$
– Thomas Andrews
May 10, 2015 at 0:44

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